Physics Test-6

By the definition of volume flow rate,

\(Q=\iint \vec{V} \cdot d \vec{s}\)

Where,

\(d \vec{s}=d \overrightarrow{s_{x}}+d \overrightarrow{s_{y}}+d \overrightarrow{s_{z}}\)

d_sx = dydz, the projection of ds on yz plane

d_sy = 0, the projection of ds on zx plane (As the surface is parallel to y axis)

ds_z = dxdy, the projection of ds on xy plane

Calculation:

To fine the relation between x, y, z, the equation of the plane ABCD,

\(\frac{x}{3}+\frac{y}{\infty}+\frac{z}{2}=1\)

\(\Rightarrow x=3-\frac{3 z}{2}\) .....(i)

⇒ Q = ∫ ∫s (x î + y ĵ). (dydz î +0+dxdy k̂)

⇒ Q = ∫ ∫s (x dydz)

As, î . ĵ = ĵ . k̂ = ĵ . î = 0 and î. î = 1

Now put the limit of y (0 to 6) and z (0 to 2) and replace the x by z using equation (i),

\(\Rightarrow Q=\int_{0}^{6} d y \int_{0}^{2}\left(3-\frac{3 z}{2}\right) d z\)

\(\Rightarrow Q=(y)_{0}^{6} \times\left(3 z-\frac{3 z^{2}}{4}\right)_{0}^{2}\)

\(\Rightarrow Q=(6) \times\left(3 \times 2-\frac{3 \times 2^{2}}{4}\right)\)

\(\Rightarrow \mathrm{Q}=18 \mathrm{~m}^{3} / \mathrm{s}\)

As we need to find out focal length of a convex lens, we will use lens maker formula, which is as below:

\(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \ldots \ldots(1)\)

Here, \(f\) is focal length of convex lens, \(\mu\) is refractive index of convex lens and \(R_1, R_2\) are radius of curvature of both surfaces of lens.As per the question, the lens is dipped in a liquid whose refractive index. So, the lens maker formula will change because in this case the refractive index of the lens will be calculated with respect to the refractive index of the liquid.

Now, \(\frac{1}{f}=\left(\frac{\mu_1}{\mu_2}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \ldots \ldots(2)\)

Here, \(\mu_1\) is refractive index of lens and \(\mu_2\) is refractive index of liquid.As given the refractive index of a lens is equal to the refractive index of liquid.

\(\mu_1=\mu_2\)

So, the ratio of both refractive index will be 1,we put it in equation (2)

\(\frac{1}{f}=(1-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

\(\Rightarrow \frac{1}{f}=0\)

\(\Rightarrow f=\infty\)

The propagation of electromagnetic waves is along the direction of cross product of electric field and magnetic field.

Electromagnetic waves are composed of oscillating magnetic and electric fields.

• The electric field and magnetic field of an electromagnetic wave are perpendicular to each other.
• These waves do not require a medium to travel. EM waves travel at a constant velocity of 3 × 10⁸ m/s in a vacuum.

Since both electric and magnetic fields are vectors, the direction of propagation of EM waves is obtained from the right-hand rule.

Let the electric field be denoted by \(\vec{E}\) and magnetic field be denoted by \(\vec{B}\).

According to the rule: If the fingers of the right hand are curled so that they follow a rotation from \(\vec{E}\) to \(\vec{B}\), then the thumb will point in the direction of the vector product i.e. the direction of EM waves.

So, the direction of EM waves is found from the cross product of the electric field and magnetic field.

\(p=\frac{C}{v^{2}}\)

\(p_{1} v_{1}^{2}=p_{2} v_{2}^{2}=C\)

\(\frac{p_{1}}{p_{2}}=\left(\frac{v_{2}}{v_{1}}\right)^{2}\)

\(v_{2}=v_{1} \sqrt{\frac{p_{1}}{p_{2}}}=0.18 \sqrt{\frac{3}{0.6}}=0.402 \mathrm{~m}^{3} / \mathrm{kg}\)

\(\mathrm{C}=\mathrm{pv}^{2}=\mathrm{p}_{1} \mathrm{v}_{1}^{2}=3 \times 0.18^{2}=0.0972 \mathrm{bar}\left(\mathrm{m}^{3} / \mathrm{kg}\right)^{2}\)

\(W=\int_{1}^{2} p d v=\int_{1}^{2} \frac{C}{v^{2}} d v=C \int_{1}^{2} \frac{d v}{v^{2}}\)

\(W=C\left[-\frac{1}{v}\right]_{v_{1}}^{v_{2}}=C\left[\frac{1}{v_{1}}-\frac{1}{v_{2}}\right]\)

\(W=0.0972 \times 10^{5}\left[\frac{1}{0.18}-\frac{1}{0.402}\right]\)

\(=29840 \mathrm{Nm} / \mathrm{kg}=29.84 \mathrm{~kJ} / \mathrm{kg}\)

Given that:

\(q_{1}=q_{2}=q \text { (let), } r_{1}=6 \mathrm{~cm}=7 \times 10^{-2} \mathrm{~m}\)

When the initial distance between them (r) = 7 cm

Then the force between them is,

\(F_{1}=K \frac{q_{1} \times q_{2}}{r_{1}^{2}}=K \frac{q \times q}{\left(6 \times 10^{-2}\right)^{2}}=\frac{K q^{2}}{36 \times 10^{-4}}\) .....(i)

When the second body is shifted away from the first by another 6 cm, then the distance between them is

r₂ = 6 + 6 = 12 cm = 12 × 10^-2 m

Then the force between them is,

\(F_{2}=K \frac{q_{1} \times q_{2}}{r^{2}}=K \frac{q \times q}{\left(12 \times 10^{-2}\right)^{2}}=\frac{K q^{2}}{144 \times 10^{-4}}\) .....(ii)

Divide equation (i) and (ii), we get

\(\frac{F_{1}}{F_{2}}=\frac{\frac{K q^{2}}{36 \times 10^{-4}}}{\frac{K q^{2}}{144 \times 10^{-4}}}=\frac{144 \times 10^{-4}}{36 \times 10^{-4}}=4\)

\(\therefore F_{2}=\frac{F_{1}}{4}\)

Given:

R₁ = 2R, R₂ = 3R, the Density of two planets is ρ₁ and ρ₂.

Acceleration due to gravity is,

\(g=\frac{G M}{R^{2}}\)

As, Mass = Volume \(\times\) Density

\(\therefore M=\frac{4}{3} \pi R^{3} \times \rho\)

So,

\(g=\frac{G\left(\frac{4}{3} \pi R^{3} \times \rho\right)}{R^{2}}\)

\(g=\frac{4}{3} \pi R G \rho\)

Since \(g, \pi, G\) is constant.

\(\therefore g \propto R \rho\)

So, acceleration due to gravity for the first planet,

\(g_{1}=R_{1} \rho_{1}\)

\(g_{1}=2 R \rho_{1}\) .....(i)

So,acceleration due to gravity for the second planet,

\(g_{2}=R_{2} \rho_{2}\)

\(g_{2}=3 R \rho_{2}\) .....(ii)

Divide equation (i) with (ii), we get

\(\frac{g_{1}}{g_{2}}=\frac{2 R \rho_{1}}{3 R \rho_{2}}=\frac{2 \rho_{1}}{3 \rho_{2}}\)

\(\therefore \frac{g_{1}}{g_{2}}=\frac{2 \rho_{1}}{3 \rho_{2}}\)

Incident wave in a string is moving from left to right (i.e. along the positive direction of x-axis) is represented by

⇒ y₁(x, t) = A sin (kx – ωt) .....(i)

Where a = amplitude and k = angular wavenumber

As there is a phase change of π radian on reflection at the fixed end of the string, therefore the reflected wave pulse travelling from right to left is represented by

⇒ y₂(x, t) = -A sin (kx + ωt) .....(ii)

According to the principal of superposition, the resultant displacement is:

⇒ y(x, t) = y₁(x, t) + y₂(x, t)

⇒ y(x, t) = A sin (kx – ωt) - A sin (kx + ωt) = -A [sin (kx + ωt) - sin (kx – ωt)]

By using the relation:\(\sin C-\sin D=2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)\), we get

⇒ y(x, t) = -(2Asin kx)cos ωt

Here, 2Asin kx = the amplitude of the wave.

\(\mu_{1}=\frac{A_{\max }-A_{\min }}{A_{\max }+A_{\min }}=\frac{2}{3}\)

\(A_{c}=\frac{A_{\max }+A_{\min }}{2}\)

\(=\frac{75+15}{2}=45\)

Now,\(P_{t 1}=P_{c 1}\left(1+\frac{\mu_{1}^{2}}{2}\right)\)

\(P_{t 2}=P_{c 2}\left(1+\frac{\mu_{2}^{2}}{2}\right)\)

The new modulated signal is given as:

\(S_{1}(t)=45\left[1+\frac{2}{3} \cos 2 \pi f_{m} t\right] \cos 2 \pi f_{c} t+A \cos 2 \pi f_{c} t\)

\(=(45+A)\left[1+\frac{30}{45+A} \cos 2 \pi f_{m} t\right] \cos 2 \pi f_{c} t\)

For this wave amplitude of the carrier is (45 + A)

88 P_t2 = 81 P_t1

\(88 P_{c 2}\left(1+\frac{\mu_{2}^{2}}{2}\right)=81 P_{c 1}\left(1+\frac{\mu_{1}^{2}}{2}\right)\)

\(88 \frac{A_{c 2}^{2}}{2}\left(1+\frac{\mu_{2}^{2}}{2}\right)=81 \frac{A_{c 1}^{2}}{2}\left(1+\frac{\mu_{1}^{2}}{2}\right)\)

Putting the values:

\(A_{c 2}=45+A\),\(A_{c 1}=45\)

\(\mu_{2}=\frac{1}{2}, \quad \mu_{1}=\frac{2}{3}\)

We get,\(A=15\)

For proper working of a clamper, the time constant of the circuit should be large because the voltage across the capacitor C does not change significantly during the time interval the diode is non-conducting.

The circuit time constant (t = RC) should be at least ten times the time period of the input signal voltage.

The clamping circuit is used when the DC value of the signal is to be shifted from one level to another level.

In a clamping circuit, the waveform of input and output remain the same, just there is a vertical shift either upward or downward.

The torque on a rectangular coil placed in a uniform magnetic field is large when the number of turns is large.

The dipole moment of the coil (M) = N I A

Torque on rectangular coil (T) = M × B = MB Sinθ = N I A B sinθ

• According to the above formula of the torque on a rectangular coil in a magnetic field, torque will be large when the number of turns is large. So, option (A) is correct and option (B) is wrong.
• If the magnetic field is perpendicular to the plane of coil then it will be along the area vector. So, angle between area vector and the magnetic field will be zero and finally the torque will be zero. So, option (C) is wrong.
• If the area will be small then torques will also be small. So, option (D) is wrong.

The intensity of light refers to the amount of photon energy per unit area.

• So, the greater the intensity of light more will be the number of photons, and as a result, more will be the number of ejected electrons. More electrons constitute more photocurrent.
• So if the intensity is doubled, the photocurrent also gets doubled.

​The photocurrent is independent of the frequency of incident light.

The energy of a photon is directly proportional to the frequency of incident light. So if the frequency is doubled, The energy is also doubled. Hence, more amount of kinetic energy is available for the emitted electrons.

​∴ The kinetic energy of the emitted electron will be increased and the photoelectric current will be 2 times.

Given that;

R = 1500 Ω

Here both resistances (each equal to R) are in parallel to each other. Also, all three resistances (each equal to R) on the right side are parallel to each other.

Use the parallel combination formula:

For the left side:

\(\frac{1}{R_{\text {left }}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}=\frac{1}{R}+\frac{1}{R}=\frac{2}{R}\)

\({R}_{\text {left }}=\frac{{R}}{2}\)

For the right side:

\(\frac{1}{R_{\text {right }}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}=\frac{3}{R}\)

\({R}_{\text {right }}=\frac{{R}}{3}\)

Now since both \(\frac{R}{2}\) and \(\frac{R}{3}\) are in parallel, so each will have the same potential difference (V = 300 Volt).

Potential difference (V) = 300 Volt

Resistance \(\frac{R}{3} = \frac{1500}{3} = 500 \Omega\)

Use Ohm’s law:

V = R I

Current in ammeter (I) \(= \frac{V}{R} = \frac{300}{500} = \frac{3}{5} A\)

A balloon moving up in the air is based on the Archimedes principle.

When the air is being heated the density of that air gets reduced compared to the density of surrounding air or we can use the gas having less density than that of air. Due to the difference in density, the balloon lifted above the ground and moves up in the air.

The point where all the masses of the system assumed to be concentrated is called center of mass of the system. For simple rigid objects with uniform density, the center of mass is located at the centroid.

For a distributed mass system the center of mass will be located at the point which is near to the mass portion having more density. In the given diagram, we have two masses- one is air and other is – sand. Air is very light compared to sand, so the center of mass will be more located in sand portion. So it will be at point D.

The torque acting on a dipole moment P placed in an electric field E is \(\vec{\tau}=\vec{p} \times \vec{E}\).

Assume an electric dipole is placed in a uniform magnetic field.

Each charge of dipole experience a force qE in electric field.

Since points of action of these forces are different, these equal and anti-parallel forces give rise to a couple that rotate the dipole and make the dipole to align in the direction of the field.

The torque τ experienced by the dipole is (qE)×(2dsinθ), where, 2d is the length of a dipole and θ is the angle between dipole and field direction.

\(\therefore\) Torque(t), \(\vec{\tau}=\vec{p} \times \vec{E}\)